graph manifolds and taut foliations - CiteSeerX

j. differential geometry 47 (1997) 446-470

GRAPH MANIFOLDS AND TAUT FOLIATIONS MARK BRITTENHAM, RAMIN NAIMI & RACHEL ROBERTS

Abstract

We examine the existence of foliations without Reeb components, taut foliations, and foliations with no S 1 S 1 -leaves, among graph manifolds. We show that each condition is strictly stronger than its predecessor(s), in the strongest possible sense; there are manifolds admitting foliations of each type which do not admit foliations of the succeeding type(s).

0. Introduction Taut foliations have been increasingly useful in understanding the topology of 3-manifolds, thanks largely to the work of David Gabai [19]. Many 3-manifolds admit taut foliations [35],[10],[30], although some do not [3],[8]. To date, however, there are no adequate necessary or sucient conditions for a manifold to admit a taut foliation. This paper seeks to add to this confusion. In this paper we study the existence of taut foliations and various re nements, among graph manifolds. What we show is that there are many graph manifolds which admit foliations that are as re ned as we choose, but which do not admit foliations admitting any further re nements. For example, we nd manifolds which admit foliations without Reeb components, but no taut foliations. We also nd manifolds admitting C (0) foliations with no compact leaves, but which do not admit any C (2) such foliations. These results point to the subtle nature behind both topological and analytical assumptions when dealing with foliations. Received February 1, 1996. The rst author was supported in part by NSF grant # DMS?9400651 and the second by an NSF Postdoctoral Fellowship. Key words and phrases. essential lamination, taut foliation, Seifert- bered space, graph manifold, Anosov ow. 446

graph manifolds and taut foliations

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A principal motivation for this work came from a particularly interesting example; the manifold M obtained by 37/2 Dehn surgery on the (?2,3,7) pretzel knot K . This manifold is a graph manifold, obtained by gluing two trefoil knot exteriors together along their boundary tori. We show that every essential lamination in M contains a torus leaf, and therefore every essential lamination intersects the image of K in M . This tells us a great deal about essential laminations in the exterior of K . This is discussed in Section 5 below. The paper is organized as follows. In Section 1 we give the necessary background on Seifert- bered spaces and graph manifolds, and introduce the appropriate numerical coordinates for describing them. In Section 2 we gather the relevant results on foliations and essential laminations to carry out our proofs. Section 3 gives the main reults of the paper, and Section 4 provides the proofs. Section 5 discusses surgery on the (?2,3,7) pretzel knot. Section 6 nishes with some speculations. This research was conducted while the authors were visiting the University of Texas at Austin in 1994-95. The authors would like to express their appreciation to the faculty and sta at that institution for their hospitality.

1. Coordinates for graph manifolds A Seifert- bered space M is an S 1-bundle whose base is a 2-orbifold. More precisely, a Seifert- bered space begins with an honest circle bundle M0 over a compact surface; for our purposes it will suce to think about a compact, orientable, surface, possibly with boundary, crossed with S 1 . To some of the boundary components of M0 we then glue a collection of solid tori, so that the meridional direction of each solid torus does not correspond to the S 1-direction on the boundary of M0 . The induced foliation of the boundaries of each of these solid tori by circles extends, in an essentially unique way, to a foliation by circles of the solid torus, so that the core of the solid torus is a leaf. This gives a foliation of M by circles, whose space of leaves - the quotient space obtained by crushing each circle leaf to a point - is a 2-orbifold. Its underlying topological space is called the base surface of the Seifert bering of M . The cone points of the orbifold correspond to the cores of the solid tori; these cores are called the multiple bers of the Seifert bering of M .Their multiplicity is the number of times nearby bers intersect a small disk transverse to the multiple ber.

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mark brittenham, ramin naimi & rachel roberts

A manifold M is a graph manifold if there is a collection T of disjoint embedded tori so that the manifold M jT obtained by splitting M open along T is a (not necessarily connected) Seifert- bered space. We assume that the collection T is minimal, in the sense that for no torus T in T is M j(T nT ) a Seifert- bered space. We adopt the convention that a Seifert- bered space is not a graph manifold, so T 6= ;. Since the bering of a Seifert- bered space is essentially unique [38], we can give a more constructive approach to minimality. Thinking in reverse, a graph manifold is obtained by gluing Seifert- bered spaces together along some of their boundary tori; the glued tori become the collection of splitting tori T . The collection T is minimal if, in gluing, the homotopy class of the circle ber in one boundary torus is not identi ed with the class of the ber in the other boundary torus. The only exceptions to this rule occur when some components are solid tori or T I ; for solid tori, minimality requires that the meridian in the boundary of the solid torus be glued to the S 1- ber, and a T I can either be absorbed into a component of M0 (if its ends are not glued together), or must have its ends glued together by a map having (on the level of H1(T ; R)) no integer-valued eigenvectors. Our results will be stated in terms of the Seifert- bered pieces making up the graph manifold, and the gluing maps between their boundary tori. To do so, we will need a proper set of coordinates. In [38] Seifert developed numerical invariants of what he called ` bered spaces', and gave a complete classi cation of them in terms of these invariants. They describe the topological type of the base orbifold, and the way that the the regular bers spin around the multiple bers. More explicitly, an orientable Seifert- bered space M can be described as follows: start with a compact surface F of genus g and b boundary components (the underlying topological space of the base orbifold), and drill out k disks (one for each multiple ber of the Seifert- bering). To be sure the resulting surface has non-empty boundary, drill out one more `zero-th' disk, giving a surface F0 . Now construct the (unique) S 1-bundle M0 over F0 with orientable total space. This bundle has a (not necessarily unique) cross-section s:F0 !M0 (because @M06= ;). The images of @F0 in @M0, together with the circle bers in @M0 , give us a system of coordinates for curves in @M0, de ning for each simple closed curve in a component of @M0 a slope in Q[f1g, where the section de nes slope 0 and the ber de nes slope 1. We then glue k + 1 solid tori back onto M0 to obtain M . The gluing of the i-th


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Figure 1

solid torus identi es the boundary of a meridian disk to some curve ai ( ber) + bi(section) in @M0. These gluings completely describe the Seifert- bered space, giving us its so-called Seifert invariant M = (g; b; a0=b0; a1=b1; : : : ; ak =bk): equals + if F is orientable, ? if not. The rational numbers ai/bi are treated as an unordered (k+1)-tuple. The denominator of each rational number (in lowest terms) turns out to be the multiplicity of the corresponding multiple ber. Since our zero-th disk did not correspond to a multiple ber, its multiplicity is 1, so b0 =1. This invariant is dependent upon the choice of section for M0; the only way this section can change, however, is by summing along vertical annuli and tori (see Figure 1). Summing along a torus does not change the associated invariant, and summing along an annulus changes the invariant in a very controlled way; it adds and subtracts 1 each from the invariants associated to the two components of @M0 containing the boundary of the annulus. We can actually remove this ambiguity by exploiting it; by a series of summings along annuli one of whose boundaries lies over the boundary of the zero-th disk, we can arrange that 0ai /bi 1, and 2[?2+ m1 ,?1] (?2,?1] (since ?( +1) can, again, be at most (m?1)/m)). Therefore, in every case, the slope of a horizontal foliation lies in (?2,0). Throwing in the possibility of a vertical sublamination which intersects the boundary adds slope 1. So to achieve our non-realizability result, we must merely construct gluing maps A:T !T so that, on the level of boundary slopes, (**) A((?2,0)[f1g)\((?2,0)[f1g)=;. This is quite readily done; for example the map A=(0,?1;1,0), which is the map A(x)=?1/x, does this. With a bit of work, it is not hard to nd many others. For A=(a; b; c; d), 1 2= A((?2,0) means ?d=c 2= (?2,0), while A(1)2= (?2,0) means a=c 2= (?2,0). Focusing on the case that Det(A)=ad? bc=1, we then have 2a ? b b A(?2,0) = 2c ? d ; d , so (**) requires (in addition to ad ? bc=1): (1) ?d=c ?2 (i.e., d=c 2) or ?d=c 0 (i.e., d=c 0),


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(2) a=c ?2 or a=c 0, and (3) b=d ?2 or (2a ? b)=(2c ? d) 0. The easiest way to arrange this is to make a; d > 0, while b; c < 0, ad ? bc = jajjdj ? jbjjcj = 1, and jaj 2jcj and jbj 2jdj. For example, A = (a; b; c; d) = (n + 1; ?(nd + d ? 1); ?1; d) with n 2 and d1, or A = (2n + 1; ?(2nm + m + n); ?2; 2m + 1) with n 2 and m 1 suce. The reader can easily supply more. The requirement 2(?2; 0) is of course necessary for the existence of a horizontal foliation, but never, in fact, sucient. Exact conditions depend upon knowing what 1 and 2 are. For example, if 1=1/3 and

2=1/5, then 0 +11/2=1=m is possible, since 1/3< 1=2 = a=m and 1/5< 1=2 = (m ? a)=m. This is in fact the best possible, since it is the largest 1/m possible, and otherwise one of 1/3,1/5 would have to be < 1=m, so m=2,3, or 4, and each case can be checked separately to see that it gives no better bounds. This analysis can be applied to any slopes 1 and 2 supplied; after nding one +1 1=m which works, one can check all smaller m's to see if a corresponding a lets +1 1=m work. Then one nds the largest m so that one of 1, 2< 1=m, and checks it and all smaller m's to see for what a's does +1 a=m work. In this way, one can nd, for example, that for

1=1/3, 2=1/5, then 2[?1; ?1=2], for

1=2/3, 2=1/5, then 2[?1; ?3=4], for

1=1/7, 2=1/5, then 2[?1; ?1=4], for

1=2/7, 2=1/5, then 2[?1; ?1=3], for

1=1/3, 2=4/5, then 2[?5=4; ?1], and for

1=2/3, 2=4/5, then 2[?3=2 ? 1;]. We know, however, from [27], that any element in the interior of such an interval can be realized by a horizontal foliation which meets the boundary in parallel loops of slope . Therefore, if the gluing map A has A(interval for rst piece) meet the interval for the second piece in its interior, then we can glue two such foliations together to obtain a taut foliation. Such a foliation usually has no compact leaves; in fact, for only one can the foliation on a Seifert- bered piece have a compact leaf (the one which sums with the i to give 0); gluing a foliation with no compact leaves to a foliation all of whose leaves meet the boundary obviously gives a foliation with no compact leaves. We also note that the generalization of Theorem B to essential laminations is true; every essential lamination in these manifolds contains

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a torus leaf. The result has the identical proof, since an essential lamination can be made transverse to the gluing torus, so that the split open pieces are essential; the split open pieces must then be horizontal, or contain a vertical annulus by Proposition 3b (again, except for (0,1;1/2,1/2)). If they are horizontal, then they extend to horizontal foliations, so their slopes fall into the same restrictive range.

Theorem C.

For this case we will use the three Seifert- bered spaces (0; 0; ?1; 1=2; 1=3; 1=6);

(0; 0; ?1; 1=2; 1=4; 1=4);

(0; 0; ?1; 1=3; 1=3; 1=3); i.e., the three Seifert- bered spaces having base S 2 with a Euclidean orbifold structure and 3 cone points. Each of these manifolds contains a horizontal torus, so can be tautly foliated by horizontal tori. By Proposition 3b, every essential lamination in these spaces is isotopic to a horizontal one. But every horizontal lamination contains a torus leaf. Matsumoto [29] outlines a proof of this in the C (2) case, using a result of Plante [33] on the polynomial growth of leaves of foliations. Plante's argument is essentially C (1), but the only place this hypothesis is used is to show that a hypothetical foliation with no compact leaves admits no null-homotopic loops transverse to the foliation. This assertion follows easily, however, either from the fact that our foliation is horizontal (transverse loops must travel non-trivially around the ber direction), or, more generally, from the C (0) proof of Novikov's theorem [39].

Theorems D and E.

In these cases we will use a graph manifold M consisting of two copies of Mi =(a once-punctured torus)S 1, i=1,2, glued together along their boundaries. They both have normalized Seifert invariant (1; 1; 0): Again, the bering on each piece is unique, so the resulting manifold is Seifert- bered if and only if the gluing map A is a shear. We therefore assume that A is not a shear. For all gluings A, the resulting manifold contains a C (0) foliation with no compact leaves. The foliation has three parts. In each piece Mi we put a vertical lamination Li =i S 1, where i is a 1-dimensional lamination in the once-punctured torus, with no compact leaves, and having


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Figure 5

Figure 6

every leaf dense in i . Such (measured) laminations are easily built carried by the standard train track in the 2-torus (Figure 5a). For every gluing A, M n(L1 [L2 ) looks essentially like (annulus)S 1 ; if we choose the standard branched surface B carrying L1 [L2 , then M0 =M nint(N(B)) has the structure of the sutured manifold (annulus)S 1 =(torus)I , with two parallel sutures on each boundary component. By foliating M0 by annuli, whose boundaries are not parallel to either of the sutures, we can, as in [19], spin the leaves in M0 along the annuli between the sutures to complete (L1 [L2 ) to a foliation (Figure 5b). The key fact in the proof of Theorem D is that no foliation of M which contains a (vertical) sublamination like one of the Li can admit a transverse C (2) structure. This is because for every (annular) leaf of the sublamination, the foliation meets the normal fence over its core in one of the patterns of Figure 6; there are closed loops limiting on on one or both sides. This implies each leaf of the sublamination has trivial linear holonomy, if the foliation has class C (2). Proposition 4 says that this is impossible, however, since the sublamination does not form an open set in M . If a C (2) foliation F of M has no compact leaves, then Proposition 1 and its extension imply that we can make F transverse to the splitting torus T , so that the induced foliations on the Seifert- bered pieces Mi are essential. Each therefore contains a vertical or horizontal sublami-

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nation, by Proposition 3. If a vertical sublamination misses T , then it comes from a 1-dimensional lamination in the interior of the base surface. It must therefore contain either a closed loop (giving a torus leaf of F ) or a lamination like the one above, so our original foliation cannot be made C (2). Hence any vertical sublamination must meet T , and the slope of the @ -foliation is 1, on one side. Since Mi does not contain a horizontal annulus, Proposition 2 implies that F cannot induce a Reeb foliated annulus on T , because it would have to be vertical when viewed from both sides, so M would be Seifert- bered. Proposition 3c and the observation above imply that if the foliation on one of the Mi meets @Mi in a foliation of slope other than 1, then the foliation is horizontal. Proposition 8 thus implies that the induced slope is in [0,1). By putting together, any C (2) foliation with no compact leaves in M can be made transverse to T ; the induced foliations on Mi meet @Mi in slopes lying in [0,1)[f1g. So to build the examples required for Theorem D, we need to nd gluing maps A=(a,b;c,d) for which A([0,1)[f1g)\([0,1)[f1g)=;. As with the proof of Theorem B, this is easily arranged. Since A(?d/c)=1, A(1)=a/c, A(0)=b/d, and A(1)=(a+b)/(c+d), after choosing det(A)=ad?bc=?1 (for convenience, so that A([0,1) will be [A(1),A(0))), we need

?d=c 62 [0; 1); a=c 62 [0; 1); and either b=d < 0 or (a + b)=(c + d) > 1: One easy way to do this is to choose c>0, d>0, a